Q.The area of the region bounded by the line $y = mx$ ($m > 0$), the curve $x^2 + y^2 = 4$ and the $x$-axis in the first quadrant is $\frac{\pi}{2}$ units. Using integration, find the value of $m$.

The key idea is to calculate the area of the region by splitting it into two parts and using definite integration.

1. Identify the boundaries and intersection point: The region is bounded by the line $y=mx$, the circle $x^2+y^2=4$ (radius $r=2$), and the $x$-axis ($y=0$) in the first quadrant. The line $y=mx$ intersects the circle at a point $(x_0, y_0)$. Substituting $y=mx$ into the circle equation gives $x^2 + (mx)^2 = 4 \implies x^2(1+m^2) = 4$. Since we are in the first quadrant, $x_0 = \frac{2}{\sqrt{1+m^2}}$.
2. Set up the definite integral: The total area $A$ can be expressed as the sum of two integrals:
   • Area under the line $y=mx$ from $x=0$ to $x=x_0$.
   • Area under the circle $y=\sqrt{4-x^2}$ from $x=x_0$ to $x=2$.

3. Evaluate the integrals:
   • The first integral: $\left[\frac{mx^2}{2}\right]_0^{\frac{2}{\sqrt{1+m^2}}} = \frac{m}{2} \left(\frac{4}{1+m^2}\right) = \frac{2m}{1+m^2}$.
   • The second integral uses the formula $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$. For $a=2$: $\left[\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right)\right]_{\frac{2}{\sqrt{1+m^2}}}^2$ $= \left(0 + 2\sin^{-1}(1)\right) - \left(\frac{1}{\sqrt{1+m^2}}\sqrt{4-\frac{4}{1+m^2}} + 2\sin^{-1}\left(\frac{1}{\sqrt{1+m^2}}\right)\right)$ $= \pi - \left(\frac{1}{\sqrt{1+m^2}}\frac{2m}{\sqrt{1+m^2}} + 2\sin^{-1}\left(\frac{1}{\sqrt{1+m^2}}\right)\right)$ …

The problem asks us to find the value of $m$ given the area of a specific region. The region is bounded by three curves: a line, a circle, and the x-axis, all in the first quadrant. Understanding the geometry of this region is crucial before setting up the integration.

The curve $x^2 + y^2 = 4$ represents a circle centered at the origin $(0,0)$ with a radius of $r=2$.

The line $y = mx$ with $m > 0$ passes through the origin and has a positive slope, meaning it lies in the first and third quadrants.

The x-axis is the line $y=0$.

The condition "in the first quadrant" restricts our attention to $x \ge 0$ and $y \ge 0$.

When a region is "bounded by the line $y=mx$, the curve $x^2+y^2=4$ and the $x$-axis" in the first quadrant, it refers to the circular sector formed by the positive x-axis, the line $y=mx$, and the arc of the circle $x^2+y^2=4$. The origin $(0,0)$ is the vertex of this sector. The radius of the sector is $r=2$. The angle of the sector is the angle that the line $y=mx$ makes with the positive x-axis. Let this angle be $\alpha$. From trigonometry, for a line $y=mx$, the slope $m$ is equal to $\tan\alpha$. Thus, $\alpha = \arctan(m)$.

The problem specifies "using integration". While the area of a sector can be found directly using the geometric formula $\frac{1}{2}r^2\alpha$, we must use integration. For a circular sector, integration in polar coordinates is the most natural and direct method.

1. Identify the boundaries in polar coordinates:

   In polar coordinates, $x = r\cos\phi$ and $y = r\sin\phi$.

   The equation of the circle $x^2+y^2=4$ becomes $(r\cos\phi)^2 + (r\sin\phi)^2 = 4$, which simplifies to $r^2(\cos^2\phi + \sin^2\phi) = 4$, so $r^2=4$. Since $r$ is a radius, $r=2$. This means the radial boundary of our region is $r=2$.

   The x-axis corresponds to the angle $\phi=0$.

   The line $y=mx$ corresponds to $\tan\phi = \frac{y}{x} = m$. So, the angular boundary is $\phi = \arctan(m)$.

   Since $m>0$, $\arctan(m)$ will be an angle in the first quadrant, which is consistent with the problem statement.

   Thus, the region is defined by $0 \le r \le 2$ and $0 \le \phi \le \arctan(m)$.

2. Set up the integral for the area:

   The area element in polar coordinates is $dA = r \, dr \, d\phi$. …